Pembahasan soal Ujian Nasional (UN) Matematika IPA jenjang pendidikan SMA untuk pokok bahasan Integral Fungsi Trigonometri.
Berikut beberapa konsep yang digunakan dalam pembahasan.
Integral Fungsi Trigonometri
1. β« sin x dx = βcos x + C
2. β« cos x dx = sin x + C
3. β« sin (ax+b) dx = \(\mathrm{-\frac{1}{a}}\)cos (ax+b) + C
4. β« cos (ax+b) dx = \(\mathrm{\frac{1}{a}}\)sin (ax+b) + C
Integral Substitusi
β« sinnax cos ax = \(\mathrm{\frac{1}{{\color{Green} a}({\color{Red} n}+1)}}\)sinn+1ax + C
β« cosnax sin ax = \(\mathrm{-\frac{1}{{\color{Green} a}({\color{Red} n}+1)}}\)cosn+1ax + C
Sudut istimewa dalam radian
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\(\frac{\pi}{6}\)
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\(\frac{\pi}{4}\)
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\(\frac{\pi}{3}\)
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|
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sin(ΞΈ)
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\(\frac{1}{2}\)
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\(\frac{1}{2}\)β2
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\(\frac{1}{2}\)β3
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cos(ΞΈ)
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\(\frac{1}{2}\)β3
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\(\frac{1}{2}\)β2
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\(\frac{1}{2}\)
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tan(ΞΈ)
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\(\frac{1}{3}\)β3
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1
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β3
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Sudut kuadrantal dalam radian
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0
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\(\frac{\pi}{2}\)
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Ο
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\(\frac{3\pi}{2}\)
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2Ο
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|
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sin(ΞΈ)
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0
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1
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0
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β1
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0
|
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cos(ΞΈ)
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1
|
0
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β1
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0
|
1
|
|
tan(ΞΈ)
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0
|
tdf
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0
|
tdf
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0
|
*tdf = tidak terdefinisi
Identitias dan sifat-sifat trigonometri
sin2x + cos2x = 1
sin2x = 1 β cos2x
cos2x = 1 β sin2x
sin 2A = 2 sin A cos A
cos 2A = cos2A β sin2A
sin2x = \(\frac{1}{2}\) β \(\frac{1}{2}\)cos 2x
cos2x = \(\frac{1}{2}\) + \(\frac{1}{2}\)cos 2x
sin A cos B = \(\frac{1}{2}\)(sin (A + B) + sin (A β B))
cos A sin B = \(\frac{1}{2}\)(sin (A + B) β sin (A β B))
cos (βx) = cos x
sin (βx) = βsin x
1. UN 2005
Hasil dari β« cos5x dx = …
A. \(\mathrm{-\frac{1}{6}}\)cos6x sin x + C
B. \(\mathrm{\frac{1}{6}}\)cos6x sin x + C
C. βsin x + \(\mathrm{\frac{2}{3}}\)sin3x
D. sin x β \(\mathrm{\frac{2}{3}}\)sin3x
E. sin x
Pembahasan :
β« cos5x dx
β β« (cos2x)2 cos x dx
β β« (1 β sin2x)2 cos x dx
Misalkan :
u = sin x
du = cosx dx
β β« (1 β u2)2 du
β β« (1 β 2u2 + u4) du
= u β \(\mathrm{\frac{2}{3}}\)u3
= sin x β \(\mathrm{\frac{2}{3}}\)sin3x
Jawaban : D
2. UN 2006
Nilai \(\mathrm{\int_{0}^{\pi }}\)sin 2x cos x dx = …
A. \(-\frac{4}{3}\)
B. \(-\frac{1}{3}\)
C. \(\frac{1}{3}\)
D. \(\frac{2}{3}\)
E. \(\frac{4}{3}\)
Pembahasan :
\(\mathrm{\int_{0}^{\pi }}\)sin 2x cos x dx
β \(\mathrm{\int_{0}^{\pi }}\)2sin x cos x cos x dx
β 2\(\mathrm{\int_{0}^{\pi }}\)cos2x sin x dx
= 2\(\mathrm{\left [ -\frac{1}{({\color{Red} 2}+1)}cos^{{\color{Red} 2}+1}x\right ]_{0}^{\pi }}\)
= \(\mathrm{-\frac{2}{3}}\)[cos3x\(]_{0}^{\pi }\)
= \(\mathrm{-\frac{2}{3}}\){cos3Ο β cos30}
= \(\mathrm{-\frac{2}{3}}\){(β1)3 β (1)3}
= \(\mathrm{\frac{4}{3}}\)
Jawaban : E
3. UN 2008
Hasil dari β« cos2x sin x dx = …
A. \(\mathrm{\frac{1}{3}}\)cos3x + C
B. \(\mathrm{-\frac{1}{3}}\)cos3x + C
C. \(\mathrm{-\frac{1}{3}}\)sin3x + C
D. \(\mathrm{\frac{1}{3}}\)sin3x + C
E. 3 sin3x + C
Pembahasan :
β« cos2x sin x dx
= \(\mathrm{-\frac{1}{({\color{Red} 2}+1)}}\)cos2+1x + C
= \(\mathrm{-\frac{1}{3}}\)cos3x + C
Jawaban : B
4. UN 2009
Hasil β« cos3x dx adalah …
A. sin x \(\mathrm{-\frac{1}{3}}\)sin3x
B. \(\mathrm{\frac{1}{4}}\)cos4x
C. 3 cos2x sin x
D. \(\mathrm{\frac{1}{3}}\)sin3x β sin x
E. sin x β 3 sin3x
Pembahasan :
β« cos3x dx
= u β \(\mathrm{\frac{1}{3}}\)u3
= sin x β \(\mathrm{\frac{1}{3}}\)sin3x
Jawaban : A
5. UN 2009
Hasil dari β« sin 3x cos x dx adalah …
A. \(\mathrm{-\frac{1}{8}}\)cos 4x \(\mathrm{-\frac{1}{4}}\)cos 2x + C
B. \(\mathrm{\frac{1}{8}}\)cos 4x
C. \(\mathrm{-\frac{1}{4}}\)cos 4x \(\mathrm{-\frac{1}{2}}\)cos 2x + C
D. \(\mathrm{\frac{1}{4}}\)cos 4x + \(\mathrm{\frac{1}{2}}\)cos 2x + C
E. β4 cos 4x β 2 sin 2x + C
Pembahasan :
β« sin 3x cos x dx
β β«\(\frac{1}{2}\)(sin (3x + x) + sin (3x β x)) dx
β \(\frac{1}{2}\) β« (sin 4x + sin 2x) dx
= \(\mathrm{\frac{1}{2}\left ( -\frac{1}{4}cos\:4x+\left ( -\frac{1}{2}cos\:2x \right ) \right )+C}\)
= \(\mathrm{-\frac{1}{8}}\)cos 4x \(\mathrm{-\frac{1}{4}}\)cos 2x + C
Jawaban : A
6. UN 2010
Hasil dari \(\mathrm{\int_{0}^{\frac{\pi }{6}}}\)(sin 3x + cos 3x) dx adalah …
A. \(\frac{2}{3}\)
B. \(\frac{1}{3}\)
C. \(0\)
D. \(-\frac{1}{3}\)
E. \(-\frac{2}{3}\)
Pembahasan :
\(\mathrm{\int_{0}^{\frac{\pi }{6}}}\)(sin 3x + cos 3x) dx
= \(\mathrm{\left [ -\frac{1}{3}cos\:3x+\frac{1}{3}sin\:3x \right ]_{0}^{\frac{\pi }{6}}}\)
= \(\mathrm{\frac{1}{3}\left [ sin\,3x-cos\,3x \right ]_{0}^{\frac{\pi }{6}}}\)
= \(\frac{1}{3}\){(sin \(\frac{\pi}{2}\) β cos \(\frac{\pi}{2}\)) β (sin 0 β cos 0)}
= \(\mathrm{\frac{1}{3}}\)((1 β 0) β (0 β 1))
= \(\mathrm{\frac{2}{3}}\)
Jawaban : A
7. UN 2010
Hasil dari β« (sin2x β cos2x) dx adalah …
A. \(\mathrm{\frac{1}{2}}\)cos 2x + C
B. β2 cos 2x + C
C. β2 sin 2x + C
D. \(\mathrm{\frac{1}{2}}\)sin 2x + C
E. \(\mathrm{-\frac{1}{2}}\)sin 2x + C
Pembahasan :
β« (sin2x β cos2x) dx
β β« β(cos2x β sin2x) dx
β β β« cos 2x dx
= \(\mathrm{-\frac{1}{2}}\)sin 2x + C
Jawaban : E
8. UN 2010
Hasil dari β« sin (\(\mathrm{\frac{1}{2}}\)x β Ο) cos (\(\mathrm{\frac{1}{2}}\)x β Ο) dx = …
A. β2 cos (x β 2Ο) + C
B. \(-\frac{1}{2}\)cos (x β 2Ο) + C
C. \(\frac{1}{2}\)cos (x β 2Ο) + C
D. cos (x β 2Ο) + C
E. 2 cos (x β 2Ο) + C
Pembahasan :
β« sin (\(\mathrm{\frac{1}{2}}\)x β Ο) cos (\(\mathrm{\frac{1}{2}}\)x β Ο) dx
β β« \(\frac{1}{2}\). 2 sin (\(\mathrm{\frac{1}{2}}\)x β Ο) cos (\(\mathrm{\frac{1}{2}}\)x β Ο) dx
β β« \(\mathrm{\frac{1}{2}}\) sin 2(\(\mathrm{\frac{1}{2}}\)x β Ο) dx
β \(\mathrm{\frac{1}{2}}\) β« sin (x β 2Ο) dx
= \(\frac{1}{2}\)(βcos (x β 2Ο)) + C
= \(-\frac{1}{2}\)cos (x β 2Ο) + C
Jawaban : B
9. UN 2011
Hasil \(\mathrm{\int_{0}^{\pi }}\)(sin 3x + cos x) dx = …
A. \(\frac{10}{3}\)
B. \(\frac{8}{3}\)
C. \(\frac{4}{3}\)
D. \(\frac{2}{3}\)
E. \(-\frac{4}{3}\)
Pembahasan :
\(\mathrm{\int_{0}^{\pi }}\)(sin 3x + cos x) dx
= \(\mathrm{\left [ -\frac{1}{3}cos\,3x+sin\,x \right ]_{0}^{\pi }}\)
= (\(-\frac{1}{3}\)cos 3Ο + sin Ο) β (\(-\frac{1}{3}\)cos 0 + sin 0)
= (\(-\frac{1}{3}\)(β1) + 0) β (\(-\frac{1}{3}\)(1) + 0)
= \(\frac{1}{3}\) + \(\frac{1}{3}\)
= \(\frac{2}{3}\)
Jawaban : D
10. UN 2011
Nilai dari β« cos42x sin 2x dx adalah…
A. \(\mathrm{-\frac{1}{10}}\)sin52x + C
B. \(\mathrm{-\frac{1}{10}}\)cos52x + C
C. \(\mathrm{-\frac{1}{5}}\)cos52x + C
D. \(\mathrm{\frac{1}{5}}\)cos52x + C
E. \(\mathrm{\frac{1}{10}}\)sin52x + C
Pembahasan :
β« cos42x sin 2x dx
= \(\mathrm{-\frac{1}{{\color{Green} 2}({\color{Red} 4}+1)}}\)cos4+12x + C
= \(\mathrm{-\frac{1}{10}}\)cos52x + C
Jawaban : B
11. UN 2012
Nilai dari \(\mathrm{\int_{0}^{\frac{\pi }{2}}}\)sin (2x β Ο) dx = …
A. β2
B. β1
C. 0
D. 2
E. 4
Pembahasan :
\(\mathrm{\int_{0}^{\frac{\pi }{2}}}\)sin (2x β Ο) dx
= \(\mathrm{\left [ -\frac{1}{2}cos(2x-\pi ) \right ]_{0}^{\frac{\pi }{2}}}\)
= \(-\frac{1}{2}\){cos (2.\(\frac{\pi}{2}\) β Ο) β cos (2.0 β Ο)}
= \(\mathrm{-\frac{1}{2}}\){cos 0 β cos (βΟ)}
= \(\mathrm{-\frac{1}{2}}\){1 β (β1)}
= β1
Jawaban : B
12. UN 2013
Nilai dari \(\mathrm{\int_{0}^{\frac{\pi }{2}}}\)sin3x dx = …
A. β1/3
B. β1/2
C. 0
D. 1/3
E. 2/3
Pembahasan :
β β« sin2x sin x dx
β β« (1 β cos2x) sin x dx
Misalkan :
u = cos x
du = βsin x dx
βdu = sin x dx
Substitusi :
β β« (1 β u2). βdu
β β« (u2 β 1) du
= \(\frac{1}{3}\)u3 β u + C
= \(\frac{1}{3}\)cos3x β cos x + C
Untuk batas x = 0 sampai x = \(\frac{\pi}{2}\)
= (\(\frac{1}{3}\)cos3\(\frac{\pi}{2}\) β cos \(\frac{\pi}{2}\)) β (\(\frac{1}{3}\)cos30 β cos 0)
= (\(\frac{1}{3}\).03 β 0) β (\(\frac{1}{3}\). 13 β 1)
= 0 β (\(β\frac{2}{3}\))
= \(\frac{2}{3}\)
Jawaban : E
13. UN 2013
Nilai \(\mathrm{\int_{0}^{\frac{\pi }{4}}}\)cos2x dx = …
B. \(\mathrm{\frac{\pi }{8}+\frac{1}{2}}\)
C. \(\mathrm{\frac{\pi }{8}-\frac{1}{4}}\)
D. \(\mathrm{\frac{\pi }{4}+\frac{1}{\sqrt{2}}}\)
E. \(\mathrm{\frac{\pi }{4}-\frac{1}{\sqrt{2}}}\)
Pembahasan :
\(\mathrm{\int_{0}^{\frac{\pi }{4}}}\)cos2x dx
β \(\mathrm{\int_{0}^{\frac{\pi }{4}}}\)(\(\frac{1}{2}\) + \(\frac{1}{2}\)cos 2x) dx
= \(\mathrm{\left [\frac{1}{2}x+\frac{1}{4}sin\,2x \right ]_{0}^{\frac{\pi }{4}}}\)
= (\(\mathrm{ \frac{\pi }{8}+\frac{1}{4} }\)sin \(\frac{\pi}{2}\)) β (0 + \(\frac{1}{4}\)sin 0)
= (\(\mathrm{ \frac{\pi }{8}+\frac{1}{4} }\).1) β (0 + \(\frac{1}{4}\).0)
= \(\mathrm{ \frac{\pi }{8}+\frac{1}{4} }\)
Jawaban : A
14. UN 2014
Hasil dari β« sin23x cos 3x dx = …
A. βsin33x + C
B. \(\mathrm{-\frac{1}{3}}\)sin33x + C
C. \(\mathrm{-\frac{1}{9}}\)sin33x + C
D. \(\mathrm{\frac{1}{9}}\)sin33x + C
E. \(\mathrm{\frac{1}{3}}\)sin33x + C
Pembahasan :
β« sin23x cos 3x dx
= \(\mathrm{\frac{1}{{\color{Green} 3}({\color{Red} 2}+1)}}\)sin2+13x + C
= \(\mathrm{\frac{1}{9}}\)sin33x + C
Jawaban : D
Hasil dari \(\mathrm{\int_{0}^{\frac{\pi }{6}}}\)sin 4x cos 2x dx = …
A. \(\frac{4}{3}\)
B. \(\frac{2}{3}\)
C. \(\frac{1}{3}\)
D. \(\frac{7}{24}\)
E. \(-\frac{1}{3}\)
Pembahasan :
\(\int_{0}^{\frac{\pi }{6}}\) sin 4x cos 2x dx
β \(\int_{0}^{\frac{\pi }{6}}\)\(\frac{1}{2}\)(sin (4x + 2x) + sin (4x β 2x)) dx
β \(\frac{1}{2}\) \(\int_{0}^{\frac{\pi }{6}}\) (sin 6x + sin 2x) dx
= \(\mathrm{\frac{1}{2}\left [ -\frac{1}{6}cos\,6x+\left ( -\frac{1}{2}cos\,2x \right ) \right ]_{0}^{\frac{\pi }{6}}}\)
= \(\mathrm{-\frac{1}{4}\left [ \frac{1}{3}cos\,6x+cos\,2x \right ]_{0}^{\frac{\pi }{6}}}\)
= \(-\frac{1}{4}\){(\(\frac{1}{3}\)cos Ο + cos \(\frac{\pi}{3}\)) β (\(\frac{1}{3}\)cos 0 + cos 0)}
= \(-\frac{1}{4}\){(\(\frac{1}{3}\)(β1) + \(\frac{1}{2}\)) β (\(\frac{1}{3}\)(1) + 1)}
= \(-\frac{1}{4}\){\(-\frac{7}{6}\)}
= \(\frac{7}{24}\)
Jawaban : D
16. UN 2014
Hasil dari β« sin3x cos x dx = …
A. \(\mathrm{\frac{1}{2}}\)sin4x + C
B. \(\mathrm{\frac{1}{4}}\)sin4x + C
C. \(\mathrm{\frac{1}{8}}\)sin4x + C
D. \(\mathrm{-\frac{1}{8}}\)sin4x + C
E. \(\mathrm{-\frac{1}{2}}\)sin4x + C
Pembahasan :
β« sin3x cos x dx
= \(\mathrm{\frac{1}{({\color{Red} 3}+1)}}\)sin3+1x + C
= \(\mathrm{\frac{1}{4}}\)sin4x + C
Jawaban : B
17. UN 2015
Hasil β« 4 sin 4x cos 2x dx adalah…
A. \(\mathrm{-\frac{1}{6}}\)cos 6x β \(\mathrm{\frac{1}{2}}\)cos 2x + C
B. \(\mathrm{-\frac{1}{3}}\)cos 6x β cos 2x + C
C. \(\mathrm{\frac{1}{6}}\)cos 6x β \(\mathrm{\frac{1}{2}}\)cos 2x + C
D. \(\mathrm{\frac{1}{6}}\)cos 6x + \(\mathrm{\frac{1}{2}}\)cos 2x + C
E. \(\mathrm{\frac{1}{3}}\)cos 6x
Pembahasan :
4 β« sin 4x cos 2x dx
β 4 β« \(\frac{1}{2}\)(sin (4x + 2x) + sin (4x β 2x)) dx
β 4.\(\frac{1}{2}\) β« (sin 6x + sin 2x) dx
= 2 (\(-\frac{1}{6}\)cos 6x + (\(-\frac{1}{2}\)cos 2x)) + C
= \(-\frac{1}{3}\)cos 6x β cos 2x + C
Jawaban : B
18. UN 2015
Nilai dari \(\mathrm{\int_{\frac{\pi }{4}}^{\pi }}\)(16 sin 2x β 2 cos 2x) dx = …
A. β9
B. β8
C. β7
D. β4
E. β2
Pembahasan :
\(\int_{\frac{\pi }{4}}^{\pi }\) (16 sin 2x β 2 cos 2x) dx
= \(\mathrm{\left [ -\frac{16}{2}cos\,2x-\frac{2}{2}sin\,2x \right ]_{\frac{\pi }{4}}^{\pi }}\)
= \(\mathrm{\left [ -8\,cos\,2x-sin\,2x \right ]_{\frac{\pi }{4}}^{\pi }}\)
= (β8cos 2Ο β sin 2Ο) β (β8cos \(\frac{\pi }{2}\) β sin \(\frac{\pi }{2}\))
= (β8(1) β 0) β (β8(0) β 1)
= β8 + 1
= β7
Jawaban : C
19. UN 2016
Hasil dari β« sin52x cos 2x dx = …
A. \(\mathrm{-\frac{1}{5}}\)sin62x + C
B. \(\mathrm{-\frac{1}{10}}\)sin62x + C
C. \(\mathrm{-\frac{1}{12}}\)sin62x + C
D. \(\mathrm{\frac{1}{12}}\)sin62x + C
E. \(\mathrm{\frac{1}{10}}\)sin62x + C
Pembahasan :
β« sin52x cos 2x dx
= \(\mathrm{\frac{1}{{\color{Green} 2}({\color{Red} 5}+1)}}\)sin5+12x + C
= \(\mathrm{\frac{1}{12}}\)sin62x + C
Jawaban : D







